Frobenius algebras and linear functionals

Posted: May 5, 2011 in Frobenius Algebras, Noncommutative Ring Theory Notes
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Throughout this post k is a field. Recall that if V is a k-vector space, then the dual space of V is the vetor spcae V^* consisting of all k-linear maps from V to k. If V is finite dimensiona, say with the basis \{v_1, \ldots , v_n \}, then for every 1 \leq i \leq n we define e_i \in V^* by e_i(v_i)=1 and e_i(v_j)=0 for all j \neq i. It is easy to see that \{e_1, \ldots , e_n \} is a basis for V^*. In particular, \dim V = \dim V^*.  

Theorem. Let A be a finite dimensional k-algebra. Then A is a Frobenius k-algebra if and only if there exists f \in A^* such that \ker f does not contain any non-zero left ideal of A.

Proof. Suppose first that A is Frobenius. So there exists a non-degenerate bilinear form B of A such that B(xy,z)=B(x,yz), for all x,y,z \in A. Define f : A \longrightarrow k by f(x)=B(x,1), for all x \in A. Obviously f \in A^* because B is linear. Now suppose that L is a left ideal of A and L \subseteq \ker f. Let y \in L and x \in A. Then xy \in L and so 0=f(xy)=B(xy,1)=B(x,y). Therefore y=0 because B is non-degenerate. So we have proved that L=0. Conversely, suppose that there exists f \in A^* such that \ker f does not contain any non-zero left ideal of A. Define B: A \times A \longrightarrow k by B(x,y)=f(xy), for all x,y \in A. Clearly B is bilinear because f is linear. Also, clearly B(xy,z)=f((xy)z)=f(x(yz))=B(x,yz), for all x,y,z \in A. To complete the proof of the theorem we only need to show that B is non-degenerate. So suppose that B(x,y)=0 for some y \in A and all x \in A. Then f(xy)=0 and thus xy \in \ker f for all x \in A, i.e. Ay \subseteq \ker f. Hence Ay=0, because Ay is a left ideal of A and \ker f does not contain any non-zero left ideal of A. So y=0 and we are done. \Box

Example 1. If K/k is a finite field extension, then K is a Frobenius k-algebra.

Proof. Let \{x_1, \ldots , x_n \} be a k-basis for K. Define f : K \longrightarrow k by f(\alpha_1 x_1 + \ldots + \alpha_n x_n)=\alpha_1, for all \alpha_i \in k. Clearly f is a non-zero k-linear map and \ker f does not contain any non-zero ideal of K because K is a field and so it has no non-zero proper ideal.

Example 2. Let G be a finite group. Then the group algebra k[G] is a Frobenius k-algebra.

Proof.  Let G=\{g_1, \ldots , g_n \}, where g_1 is the identity element of G. Define the map f : k[G] \longrightarrow k by f(\alpha_1g_1 + \ldots + \alpha_n g_n)=\alpha_1, for all \alpha_i \in k. Clearly f is a k-linear map and \ker f =kg_2 + \ldots + kg_n. Suppose that L \subseteq \ker f and L is a non-zero left ideal of k[G]. Let 0 \neq x= \sum_{i=2}^n \alpha_ig_i \in L. Then \alpha_j \neq 0, for some j \geq 2. Since L is a left ideal of k[G], we have g_j^{-1}L \subseteq L \subseteq \ker f. But the coefficient of g_1 in g_j^{-1}x is \alpha_j \neq 0 and so g_j^{-1}x \notin \ker f, which is a contradiction. Thus L=0. \ \Box     

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